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ella [17]
3 years ago
11

Which of the following is not true in regard to noble gases? A. These elements have full outermost shells. B. These elements are

found in the last column (Group 18) on the periodic table. C. These elements tend to be relatively inactive. D. These elements tend to gain or lose electrons easily
Chemistry
2 answers:
Pachacha [2.7K]3 years ago
8 0
<h2>Answer:</h2>

D. These elements tend to gain or lose electrons easily

<h2>Explanation:</h2>

The gases which are considered as Nobel are generally nonreactive. Truth be told, they are the least responsive components on the periodic table. This is on the grounds that they have a total valence shell. They tend to pick up or lose electrons rarely.

In 1898, Hugo Erdmann begat the expression “Nobel gas” to mirror the low reactivity of these components, similarly as the respectable metals are less receptive than different metals. These gases have high energies of ionization and immaterial electronegativities. These gases have low breaking points and are on the whole gases at room temperature.

Likurg_2 [28]3 years ago
5 0

Answer:

D.These elements tend to gain or lose electrons easily is <em>incorrect.</em>

Explanation:

Noble gasses consist of atoms that have full rings of electrons. Electrons are what bond atoms together to create chemical compounds, and incomplete rings are needed for the process.

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A laser produces 17.0 mW of light. In 4.00 hr , the laser emits 6.04×1020 photons. What is the wavelength of the laser?
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The unit 'mW' means milliwatts. It is a unit of work. There are 1,000 milliwatts in a 1 Watt of work. In 4 hours, there are 14,400 seconds.

Work= Energy/time
17 mW * 1 W/1000 mW = Energy/(14,400 seconds)
Solving for energy,
Energy = 244.8 J
Energy/photon = 244.8 J/(6.04×10²⁰) = 4.053×10⁻¹⁹ J/photon

Using the Planck's equation:

E = hc/λ
where h = 6.626×10⁻³⁴ m²·kg/s, c = 3,00,000,000 m/s and λ is the wavelength

4.053×10⁻¹⁹ J/photon = (6.626×10⁻³⁴ m²·kg/s)(3,00,000,000 m/s)/λ
λ = 4.9×10⁻⁷ m or 49 micrometers


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Answer: can someone please translate so i can answer.

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